Using the digits 2, 3, 4, 7 and 8, Carlos will form five-digit positive integers. Only the digit 2 can be used more than once in any of Carlos’ five-digit integers. How many distinct fivedigit positive integers are possible?

Guest May 29, 2020

#1**+3 **

_{Cases:}

_{No repetition of 2:}

\(cases = {5!}\)

_{2 is present two times:}

\(cases = {{5!} \over 2!}\)

_{2 is present three time:}

\(cases = {{5!} \over 3!}\)

_{2 is present four time:}

\(cases = {{5!} \over 4!}\)

_{2 is present five time:}

\(cases = {{5!} \over 5!}\)

_{all cases:}

\(total = {5!+{{5!} \over 2!}+{{5!} \over 3!}+{{5!} \over 4!}+{{5!} \over 5!}}\)

\(total = {206}\)

Hephaestus May 29, 2020

#2**0 **

A computer code calculates a total of 501 distinct permutations and begins with:

{{2, 2, 2, 2, 2}, {2, 2, 2, 2, 3}, {2, 2, 2, 2, 4}, {2, 2, 2, 2, 7}, {2, 2, 2, 2, 8}, {2, 2, 2, 3, 2}, {2, 2, 2, 3, 4}, {2, 2, 2, 3, 7}, {2, 2, 2, 3, 8}, {2, 2, 2, 4, 2}, {2, 2, 2, 4, 3}, {2, 2, 2, 4, 7}, {2, 2, 2, 4, 8}, {2, 2, 2, 7, 2}, ...........and ends with:

{8, 4, 7, 3, 2}, {8, 7, 2, 2, 2}, {8, 7, 2, 2, 3}, {8, 7, 2, 2, 4}, {8, 7, 2, 3, 2}, {8, 7, 2, 3, 4}, {8, 7, 2, 4, 2}, {8, 7, 2, 4, 3}, {8, 7, 3, 2, 2}, {8, 7, 3, 2, 4}, {8, 7, 3, 4, 2}, {8, 7, 4, 2, 2}, {8, 7, 4, 2, 3}, {8, 7, 4, 3, 2} = 501 permutations.

Guest May 29, 2020

#3**0 **

OK, young person! If you don't believe it, here are the 16 distinct COMBINATIONS, which you must permute as follows:

[(2, 2, 2, 2, 2), (2, 2, 2, 2, 3), (2, 2, 2, 2, 4), (2, 2, 2, 2, 7), (2, 2, 2, 2, 8), (2, 2, 2, 3, 4), (2, 2, 2, 3, 7), (2, 2, 2, 3, 8), (2, 2, 2, 4, 7), (2, 2, 2, 4, 8), (2, 2, 2, 7, 8), (2, 2, 3, 4, 7), (2, 2, 3, 4, 8), (2, 2, 3, 7, 8), (2, 2, 4, 7, 8), (2, 3, 4, 7, 8)] >>Total = 16 distinct combinations.

[1 + 5*4 + 6*20 + 4*60 + 1*120] =501 distinct permutations.

Guest May 29, 2020